Question

If $(10{)}^9+2(11{)}^1(10{)}^8+3(11{)}^2(10{)}^7+...+10(11{)}^9=k(10{)}^9,$ then k is equals to:
(IIT JEE Main 2014)

• A. $\frac{121}{10}$
• B. $\frac{441}{100}$
• C. 100
• D. 110

Answer 1

The correct option is C

100

We have,
$(10{)}^9+2(11{)}^1(10{)}^8+3(11{)}^2(10{)}^7+...+10(11{)}^9-k(10{)}^9$
Dividing by $(10{)}^9$ on both sides, we get
$1+2\left(\begin{array}{c}\frac{11}{10}\end{array}\right)+3{\left(\begin{array}{c}\frac{11}{10}\end{array}\right)}^2+...+10{\left(\begin{array}{c}\frac{11}{10}\end{array}\right)}^9=k...\left(i\right)$
Multiplying $\left(\begin{array}{c}\frac{11}{10}\end{array}\right)$on both side of above equation we get,
$\left(\begin{array}{c}\frac{11}{10}\end{array}\right)+2{\left(\begin{array}{c}\frac{11}{10}\end{array}\right)}^2+3{\left(\begin{array}{c}\frac{11}{10}\end{array}\right)}^3+...+10{\left(\begin{array}{c}\frac{11}{10}\end{array}\right)}^{10}=\left(\begin{array}{c}\frac{11}{10}\end{array}\right)k...\left(ii\right)$

Subtracting (ii) from (i), we get
$k-\left(\begin{array}{c}\frac{11}{10}\end{array}\right)k=1+\left(\begin{array}{c}\frac{11}{10}\end{array}\right)+{\left(\begin{array}{c}\frac{11}{10}\end{array}\right)}^2+...+{\left(\begin{array}{c}\frac{11}{10}\end{array}\right)}^9-10{\left(\begin{array}{c}\frac{11}{10}\end{array}\right)}^{10}$

$\Rightarrow k\left(\begin{array}{c}1-\left(\begin{array}{c}\frac{11}{10}\end{array}\right)\end{array}\right)=\frac{1\left(\begin{array}{c}{\left(\begin{array}{c}\frac{11}{10}\end{array}\right)}^{10}-1\end{array}\right)}{\left(\begin{array}{c}\frac{11}{10}\end{array}\right)-1}-10{\left(\begin{array}{c}\frac{11}{10}\end{array}\right)}^{10}$

$\Rightarrow k\left(\begin{array}{c}1-\left(\begin{array}{c}\frac{11}{10}\end{array}\right)\end{array}\right)=-10$

$\Rightarrow k=100$